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PROFESSOR: Hi.
Today we're going to conclude our study of vectors
as applied to motion in the plane.
Now recall that for the last two units,
we were discussing polar coordinates.
So today what we would like to do
is investigate what velocity and acceleration vectors would
have looked like, had we elected to pick representative vectors
in terms of polar coordinates.
Now what do we mean by representative vectors?
We mean, of course, analogs of i and j,
just like T and N in tangential-normal components
were parallels to i and j.
Let's take a look and see what that means here.
We call today's lecture "Vectors in Polar Coordinates."
The idea is that we're given a curve
C, which we have for some reason or other
elected to express in terms of polar coordinates.
The polar equation of the curve C is r equals f of theta.
A typical point on the curve C would
be called r comma theta say, where theta was the angle made
by the horizontal and the radius vector,
and r was the distance from the origin to the point.
Now what we're saying is, that in terms
of polar coordinates, a very natural vector
to pick-- especially if we think later
in terms of the simple force fields
that we've talked about earlier, the idea
that there may be a force that has its line of action
from the origin to the point on the curve--
a very natural vector to choose is the vector
that we elect to call u sub r.
And what that vector is, apparently,
is the vector 1 unit long having the direction
of the radius vector R. So here is u sub r over here.
If we think of u sub r as playing the role of i,
then the vector which plays the role of j
should be a positive 90-degree rotation of u sub r.
And we elect to call that vector u sub theta, using the theta
here more to indicate the fact that we're
using polar coordinates than to indicate anything
about the angle theta itself.
In other words, notice that u sub theta, by definition,
is just a positive 90-degree rotation of u sub r, where
u sub r is a unit vector in the direction of the radius vector.
Now if we want to see this in terms of i
and j components, what we're saying
is that u sub r is a unit vector whose i component is what?
Since this angle here is also theta,
its i component is cosine theta, and its j component
is sine theta
So u sub r is cosine theta i plus sine theta j.
u sub theta-- and I'm going to do a little twist here
that I didn't do in the T and N components,
just to show you another approach.
Rather than to start with derivatives or anything
like this, notice that what we know about u sub theta
is that it's obtained from u sub r
by a positive 90-degree rotation of theta.
So that means if I replace theta in the expression for u sub r,
by theta plus 90 degrees, that should give me u sub theta.
If I now remember my trigonometric identities,
that tells me that u sub theta is minus sine theta
i plus cosine theta j.
By the way, if we now look at this expression
and compare it with the expression for u sub r,
we see at once that u sub theta is the derivative of u sub
r with respect to theta.
And notice, as we said before, that part of this
should have been known to us by now.
Namely, since u sub r varies with theta
but it has a constant magnitude, we
know that the derivative of u sub r with respect to theta
has to be perpendicular to u sub r.
In other words, we knew that u sub theta had
to be either plus or minus the derivative of u sub
r with respect to theta, but now we
have a direct way of showing this.
And by the way, going one step further,
if we now differentiate u sub theta with respect to theta,
we get what?
Minus cosine theta i minus sine theta j,
which is just minus u sub r.
In other words, if you differentiate
u sub r with respect to theta once,
you get u sub theta, just as we should.
If you differentiate a second time, you get minus u sub r.
And therefore, it appears that the operation
of differentiating with respect to theta rotates
u sub r by 90 degrees.
By the way, I should mention that I
could have made all of these remarks
when we were studying tangential and normal components.
In other words-- I just wrote this out here,
but this was true with T and N. But the point was,
we never had to differentiate N with respect
to t to find the acceleration vector a.
In other words, recall that the key step in using
tangential and normal vectors-- and I'll mention this
in a little bit more detail later--
was that the velocity vector was simply
ds/dt times the unit tangent vector T. The coefficient of N
was 0.
In polar coordinates, notice that v in general
will have both a u sub r and a u sub theta component.
Therefore, to compute a, I have to differentiate v with respect
to t.
That means, among other things, I'm
going to have to take the derivative of u sub
theta with respect to t.
By the chain rule, that's going to be
the same as taking the derivative of u sub
theta with respect to theta times d theta / dt.
But the important point is, is that someplace
along the line in studying kinematics
and polar coordinates, I am going
to have to differentiate u sub theta with respect to theta.
And just to show you again very, very quickly
what I mean by this, all I'm saying
is we already know in kinematics that the velocity vector is
always tangential to the curve.
Notice in this particular diagram,
for example, that if you look at u sub r and u sub theta,
if you think of a vector whose direction is
tangent to the curve at this particular point,
that vector will have, in general,
both a u sub theta and a u sub r component.
In fact, in this particular diagram,
I shouldn't say "in general," it will have u sub r and a u sub
theta component.
OK.
So far so good, but now I want to make one little caution,
a caution which is not at all self-evident, at least to me,
and which gave me great difficulty myself
when I was a student.
And that is, my feeling was that u sub r was simply the unit
vector in the direction of r.
In fact, I said that earlier, that u sub
r was the unit vector in the direction of the radius vector
R. And this is one of the reasons why even though
Professor Thomas in the textbook doesn't make such
an issue over this, why I am such a bug on using the phrase
"sense" as well as direction.
And that is, my claim is that the unit vector
u sub r need not be the radius vector R divided
by the magnitude of R. It'll have the same direction,
but watch what happens with sense.
Instead of talking about this thing abstractly,
let me give you a concrete example.
Let's take the curve which in polar coordinates
has the equation r equals cosine theta.
OK?
As you recall, this would be this particular circle here.
Now let's take theta to be 120 degrees.
If I use our definition for u sub r, which is cosine theta
i plus sine theta j, and replace theta
by 120 degrees, what I get is, is that u sub r
is cos 120 degrees i plus sine 120 degrees j.
Remember that the cosine of 120 is minus 1/2
and the sine of 120 is plus 1/2 the square root of 3.
u sub r turns out to be minus 1/2 i plus 1/2
the square root of 3 j.
On the other hand, my claim is that when theta is 120 degrees,
what point are we at the curve?
You see, if I take theta to be 120
degrees-- notice that when theta is 120 degrees,
r is negative 1/2.
And that therefore I'm at the point P_0 here.
Recall that by definition, R, the radius vector,
is measured from the origin to the point.
In other words, according to our previous definition,
it's this vector, which would be called
R. But our definition says that it's
this vector, which is u sub r.
And in fact, if you just check the figures
that we've obtained over here, notice
that u sub r has its i component equal to minus 1/2,
its j component being plus 1/2 the square root of 3.
Therefore x is negative.
y is positive.
But if x is negative and y is positive,
you're in the second quadrant, not the fourth quadrant.
You see?
In other words, u sub r is almost the radius vector.
In fact, it would have been, if in polar coordinates,
little r happened to be positive.
In fact, let me summarize that in a different way.
Let's assume that we have a curve C whose polar equation
is r equals f of theta.
Then the idea is this.
If f of theta happens to be at least as big as 0,
then u sub r is equal to the radius vector
R divided by its magnitude.
In other words, u sub r will be in the same direction
as the radius vector.
And by the way, recall this is just another way of saying r.
What we're saying is again, if we had never
let little r in polar coordinates be negative,
no problem would have occurred.
But we do let little r be negative.
So we have to be a mite careful.
The careful part comes in where?
If r happens to be negative.
In which case, u sub r has the opposite sense
of the radius vector capital R, which
we saw in the previous example.
In other words, in this case, u sub r
is the negative of the radius vector
divided by its magnitude.
In what case is that?
If r happens to be negative.
The important point to notice, however, that in either case,
the radius vector R is equal to the polar coordinate r times u
sub r.
In other words, if r happens to be positive,
these two vectors have the same sense.
If r happens to be negative, these two vectors
have the opposite sense.
In other words, in either case, this expression
is always correct.
But the important thing to notice
is that the sense of u sub r is determined by theta, not
by r, not by f of theta.
OK?
At any rate, once we have this particular recipe established,
we can now go ahead and study motion in the plane.
Namely, notice that our radius vector
R is now given by the polar coordinate r times u sub r.
Or I guess I should say here that I'm
assuming that the equation of motion
is given by r is some function of theta.
That's the r that I'm using in here.
At any rate, what is the velocity vector?
By definition, it's just the derivative of the radius vector
with respect to time.
That's just d/dt of r times u sub r.
Now keep in mind, that r and u sub r
are both functions of time-- namely,
the distance of the particle from the origin
as well as the direction of the line of action
that joins the particle to the origin
will, in general, depend on time.
Consequently, I must use the product rule here.
I already know that I can use the product rule for vector
and/or scalar functions and any combination thereof.
So I just differentiate this thing with respect to time.
I get what?
This is dr/dt-- in other words, the derivative
of the first times the second, which is u sub r, OK?
Plus the first times the derivative
of the second, which is the derivative of u
sub r with respect to time.
Now keep in mind, again, there is
nothing wrong with this recipe.
But what I would like is to have the velocity expressed in u sub
r and u sub theta components.
So far, I have it expressed in terms of u sub r component
and a d(u sub r)/dt component.
But now, here again is why the chain rule is so important.
Keep in mind that I already know that if this expression here
had been the derivative of u sub r with respect
to theta instead of with respect to t,
what would this expression have been?
It would have been u sub theta.
We solved that earlier in the lecture.
Well, here's what we do.
We say, OK, these aren't the same.
But let's cross this out.
Let's use the chain rule.
And by the chain rule, the derivative
of u sub r with respect to t is the same as the derivative
of u sub r with respect to theta-- times d theta / dt.
And rewriting this so that it becomes legible,
we have that the velocity vector is dr/dt times u sub r plus r d
theta / dt times u sub theta.
And again notice that as far as u sub r and u sub
theta are concerned, even if I did not notice my subtlety--
and by the way, I'm going to leave this for the exercises--
but even if I didn't notice the subtlety that u sub r need not
have the same sense as the radius vector r-- notice
that if I did not try to draw this thing to scale,
I can still get the-- I shouldn't have said the scale--
but if I didn't try to graph the answer here given
r as a function of theta and theta is a function of t,
notice that dr/dt and r d theta / dt
are well-defined arithmetically with no possible chance
of making a geometrical mistake.
The place that you can make the biggest mistake
is if you automatically think that u sub r must
have the same sense as capital R. But as I say,
we'll leave any additional discussion
of that for the exercises.
I should also point out that when I first
learned this recipe myself, it turned out
that we were ahead of-- the physics class was ahead
of the math class.
And we learned this thing in the physics class
almost intuitively.
In other words, as a geometric aside,
notice that if I'm given the curve
and say s indicates the direction of increasing arc
length here, what I could do is think of a little differential
region here.
Namely, here's my radius vector r,
and here's my velocity vector in the direction of u sub r.
Then I take a little increment of angle d theta,
and I now think of v sub theta, which is at right angles
to v sub r, as being tangent to the circle I would have
obtained if I had imagined that this particular point--
the particle was being viewed with respect to the circle
rather than to the curve itself.
To make a long story short, what I'm driving at
is that physically, it's very easy
to justify that the magnitude of the u sub r
component of the velocity is the magnitude of dr/dt--
how fast the radius vector is changing instantaneously.
On the other hand, notice that for the u sub theta component,
this arc length is given in differential form by r d theta.
If I divide the arc length by the time, which is dt,
I get r d theta divided by dt, which leads to r d theta / dt,
which is the same expression that we got analytically.
But the point that I want to bring out here
is that our derivation required no geometrical physical
insight.
And the reason that I want to bring this out
is I followed this argument fine in my elementary physics
course.
The place I got hung up is that the instructor then went
into a fantastic hand-waving type of demonstration
and showed us how the acceleration looked
in terms of u sub r and u sub theta components.
And actually, that was a blessing for me,
because it was that day that I decided
to become a math major rather than a physics
major, which was a blessing both for me and society, I guess.
But the thing that I want to show you
is that the beauty of our mathematical approach
is that we can now obtain a, the acceleration vector,
from the velocity vector without having to know
any great physical insight.
In fact, we have to know no physical insight to do this.
Namely, by definition, a is the derivative of the velocity
vector with respect to time.
We also have seen that the velocity
vector is the expression that I have here in brackets.
So I have to differentiate that with respect to time.
Notice that this is the sum of two terms, one of which
is a product of two factors, and the other of which
is a product of three factors.
And by the way, among other things to review here,
this is the first time in this course
that we have actually had to use the product
rule for a function consisting of three variable factors.
Even though we discussed this in part one,
here is a case where in a real-life situation,
what we need is the derivative rule for a product of three
functions.
At any rate, this is done in great detail in the text.
I do it more in the notes.
So I'm just going to hit the highlights here.
The point is I now differentiate this sum term by term.
Namely, to differentiate this, I take
the derivative of the first term times the second
plus the first term times the derivative of the second.
Now to differentiate this term, I
have to differentiate a product of three factors.
And recall-- and by the way, as I told you in part one,
whenever I say "recall," that means if you don't recall,
it's my polite way of saying, "look it up."
But to differentiate a product of three functions,
we write the product down three times,
and each time differentiate a different factor.
For example, the first time we'll
differentiate r with respect to t, which is dr/dt.
The second time, we'll differentiate
d theta / dt with respect to t, which
is the second derivative of theta with respect to t.
And the third time, we'll differentiate
u sub theta with respect to t, which is the derivative of u
sub theta with respect to t.
And summarizing that, what do I have here?
I have dr/dt d theta / dt times u sub theta plus r d^2 theta /
dt squared times u sub theta plus r d theta / dt times
the derivative of u sub theta with respect to t.
Now the point is that if I look at these five terms,
some of them are in nice form.
Namely, here's a u sub r term.
Here's a u sub theta term.
But these terms are sort of mongrelized.
Namely, what I have to do here is utilize the chain rule.
And remember that the derivative of u sub
r with respect to theta would have been u sub theta.
The derivative of u sub theta with respect to theta
would have been minus u sub r.
So by the chain rule-- you see what
I'm going to do is, I'll replace each of these terms
by their chain rule expression.
Then I'll collect terms.
And the reason I'm going over this
fairly rapidly is that it is a problem of sheer mechanics.
But the punch line is that if I now collect my terms,
the acceleration vector has as its u sub r component d^2 r /
dt squared minus r d theta / dt squared.
And the u sub theta component is r d^2 theta / dt squared plus 2
dr/dt d theta / dt.
And the beautiful part, from my point
of view about all of this, is if I
don't understand any physics at all,
this particular result is valid.
It's mathematically self-contained.
Now certainly there is no harm in a man
who understands physics well enough to say, look at it.
This is the acceleration in the radius direction alone.
And this is some kind of a correction factor proportional
to the square of the angular velocity, see,
d theta / dt being angular velocity and what have you.
And go through this particular thing.
I'm saying fine, if you can do that.
But notice the beauty.
This complicated expression gives us
the acceleration vector in terms of u sub r and u sub
theta with no hand-waving.
It's mathematically self-contained.
And by the way, keep in mind that one
of the reasons that we study polar coordinate motion
is the fact that, in many cases, we
are going to be dealing with a central force field.
And the interesting thing is that in a central--
I'll just abbreviate this-- in a central force situation,
this expression is 0.
See, central force means what?
That the force is in the radial direction.
That means all of the acceleration--
if you're using Newtonian physics,
F equals ma-- all the acceleration is
in the direction of u sub r.
Therefore, the component in the direction of u sub theta
must be 0.
So this fairly complicated expression--
r d^2 theta / dt squared plus 2 dr/dt d theta / dt equals 0
becomes the fundamental equation for central force field motion.
But again, we'll talk about that more in the exercises.
What I wanted to do now was to make what I think
is a very important summary.
And that is that when we're studying the position vector
R, and the velocity vector v, and the acceleration vector a,
that none of these depend on the coordinate system.
It's only their components that do.
In other words, at the expense of having
a fairly jumbled figure which I rationalize here-- it is small,
but I think it is clear from context.
What I'm saying is, let's suppose I have a curve C,
and some point P_0 on this curve C.
I can draw in the pair of orthogonal vectors
i and j in the plane.
I can draw in the pair of orthogonal vectors
u sub r-- "orthogonal" means perpendicular,
if we haven't said that before-- u sub r and u sub theta.
I can draw those in.
And I can draw in T and N. Now all
I know is that if I have the velocity vector v,
it must be tangential to the curve.
Hopefully by this time, we realize that the acceleration
vector has no such restriction.
Let's just draw in a v and an a, call these the velocity vectors
and the acceleration vectors.
The point is that v and a are determined by the motion--
not by the coordinate system.
In other words, when we're talking about the velocity
of this particle at the point P:0, its velocity is the same,
no matter what coordinate system we're using.
It just happens that if we're dealing
with Cartesian coordinates, the velocity vector
is dx/dt i plus dy/dt j.
In other words, it's this particular combination
of i and j.
If we're using T and N components,
the particular combination of T and N is what?
ds/dt times the unit tangent vector plus 0 N.
And if we happen to be using polar coordinates,
the expression is dr/dt * u sub r plus r * d theta / dt
* u sub theta.
But let me circle these, because it's
the same velocity in each case.
We use this when horizontal and vertical motion are important.
We use this when we're interested in motion
along the curve.
And we use this primarily in central force fields.
But it makes no difference.
It's the same velocity vector.
And in a similar way, it's also the same acceleration vector,
whichever system you happen to use.
Namely, if we use Cartesian coordinates,
the acceleration vector is the second derivative
of x with respect to t times i plus the second derivative of y
with respect to t times j.
That same vector, if we express it in T and N components,
is d^2 s / dt squared times T, plus kappa,
the curvature number, times ds/dt squared times N.
And if we express it in terms of polar coordinates,
as we just saw earlier in our lecture,
this is the expression that we get.
In other words then, this summarizes our study
of motion in the plane using either Cartesian
or polar or tangential and normal components.
You see, the point is that we pick whichever
coordinate system happens to be of the greatest interest to us,
the greatest value to us.
We make the coordinate system our slave,
rather than the other way around,
and tackle the problem from that particular point of view.
At any rate, that ends this phase of our particular course.
And in the next phase of our course,
we get to probably what is the most fundamental building
block of the entire course.
We get to that particular topic which,
by and large, most courses in functions of several variables
begin with.
But we'll talk about that more the next time we meet.
And until that time, goodbye.
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